PAT顶级题解目录
Here is a simple intersting letter-moving game. The game starts with 2 strings S and T consist of lower case English letters. S and T contain the same letters but the orders might be different. In other words S can be obtained by shuffling letters in String T. At each step, you can move one arbitrary letter in S either to the beginning or to the end of it. How many steps at least to change S into T?
Input Specification:
Each input file contains one test case. For each case, the first line contains the string S, and the second line contains the string T. They consist of only the lower case English letters and S can be obtained by shuffling T's letters. The length of S is no larger than 1000.
Output Specification:
For each case, print in a line the least number of steps to change S into T in the game.
Sample Input:
iononmrogdg
goodmorning
Sample Output:
8
Sample Solution:
(0) starts from iononmrogdg
(1) Move the last g to the beginning: giononmrogd
(2) Move m to the end: giononrogdm
(3) Move the first o to the end: ginonrogdmo
(4) Move r to the end: ginonogdmor
(5) Move the first n to the end: gionogdmorn
(6) Move i to the end: gonogdmorni
(7) Move the first n to the end: googdmornin
(8) Move the second g to the end: goodmorning
题目大意：给出两个序列，保证两个序列的组成（字符）相同，现在问要将序列2变为序列1需要几步？
解题思路：其实题目的意思就是求两个字符串的最长公共子序列。我们要将word2恢复到word1的状态，要找到word2与word1最长公共子序列，可以用暴力搜索，最后用word1的长度减去公共子序列的长度就可以了。
代码：
#include<bits/stdc++.h>
using namespace std;
int main(){
string word1, word2;
cin >> word1 >> word2;
int m = 0;
for(int i = 0; i < word2.length(); ++ i){
int k = 0, j = i;
for(char c : word1)
if(word2[j] == c){
++ j;
++ k;
}
m = max(m, k);
}
printf("%d", word1.length()-m);
}